Stability constants, Kstab

· Kstab = the equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules.
· It measures how far the equilibrium lies towards the complex ion product.
· Large Kstab = equilibrium lies far to the right = more stable complex ion.
· Small Kstab = equilibrium lies less far to the right = less stable complex ion.
· For CIE, remember: Kstab is used for complex formation equilibria, especially in ligand exchange questions.

This image shows ligand exchange in copper(II) complexes. It helps connect complex formation with the colour changes often seen when more stable complexes form. Source

Writing Kstab expressions

· Write the balanced equilibrium equation for formation of the complex.
· Put the complex ion concentration on top.
· Put the reactant ion/ligand concentrations on the bottom, each raised to its stoichiometric coefficient.
· Do not include [H₂O] in the expression because water is the solvent and its concentration is effectively constant.
· General form: M + nL ⇌ MLₙ
· Kstab = [MLₙ] / ([M][L]ⁿ)
· Example:
[Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O
Kstab = [[Cu(NH₃)₄(H₂O)₂]²⁺] / ([[Cu(H₂O)₆]²⁺][NH₃]⁴)
· Exam trap: square brackets around complexes can mean formula brackets or concentration brackets; in Kstab expressions, brackets mean concentration.

Using Kstab in calculations

· Use equilibrium concentrations, not initial concentrations, in the final Kstab expression.
· If the question gives moles and volumes, first calculate concentrations using c = n/V.
· If solutions are mixed, use the total final volume after mixing.
· Rearrange the expression carefully to find the unknown concentration.
· For MLₙ, remember the ligand concentration is raised to n, e.g. [NH₃]⁴ for four NH₃ ligands.
· A very large Kstab often means almost all metal ion has formed the complex, but only make approximations if the question allows or the data clearly supports it.
· Keep answers to an appropriate number of significant figures.

Ligand exchange and Kstab

· Ligand exchange happens when one ligand in a complex is replaced by another ligand.
· The complex with the larger Kstab is the more stable complex.
· In a competition between ligands, the ligand forming the complex with the larger Kstab is more likely to replace the other ligand.
· A ligand exchange equilibrium tends to move towards the side containing the more stable complex ion.
· CIE wording to use: “The larger Kstab shows that the complex ion formed is more stable, so the equilibrium lies further to the right.”
· When comparing complexes, use Kstab values, not just ligand charge or ligand size, unless the question asks for a qualitative explanation.

Interpreting large Kstab values

· Large Kstab = products strongly favoured = stable complex ion.
· A stable complex has a low tendency to dissociate back into separate metal ions and ligands.
· If two complexes are compared:
· higher Kstab → more stable complex
· lower Kstab → less stable complex
· In ligand exchange, the complex with the greater Kstab forms preferentially.
· Be careful: Kstab does not measure rate; it measures equilibrium position/stability.

These structures show how multidentate ligands can form very stable complexes. They are useful for linking Kstab values with ligand exchange and complex stability. Source

Exam method for Kstab questions

· Step 1: Write the balanced complex formation equilibrium.
· Step 2: Write Kstab = products / reactants using concentrations.
· Step 3: Omit H₂O from the expression.
· Step 4: Substitute equilibrium concentrations into the expression.
· Step 5: Rearrange and calculate the unknown, checking powers such as [ligand]², [ligand]⁴ or [ligand]⁶.
· Step 6: Interpret the value: larger Kstab = more stable complex ion.